\(\Leftrightarrow\dfrac{1}{2}x+\dfrac{2}{3}x-x=-4\Leftrightarrow\dfrac{3x+4x-6x}{6}=-\dfrac{24}{6}\)

\(\Rightarrow x=-24\)

b) \(x.\left(\dfrac{1}{2}+\dfrac{2}{3}-1\right)=-4\)

\(x.\dfrac{-1}{6}=-4\)

\(x=-4:\dfrac{-1}{6}\)

\(x=-24\)

Ta có: \(0.5x+\dfrac{2}{3}x-x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{6}=-4\)

hay x=-24

b) 8

c) 3

b) 50-3(x+4)=14

3(x+4)=36

x+4=13

x=9

c)2⁸‐ⁿ+75=107

2⁸-ⁿ=32

2⁸-ⁿ=2⁵

8-x=5

x=3

\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

b ) - 25 + ( - 16 + x ) = 0

               ( - 16 + x ) = 0 - ( - 25 )

                 - 16 + x   = 25

                          x    = 25 - ( - 16 )

                          x    = 41

Vậy x = 41

Có ai ko guips mk vs

x=7,y=1

_HT_

\(B=\overline{2x10y9}⋮9\left(0\le x,y\le9\right)\)

\(\Rightarrow\left(2+x+1+0+y+9\right)⋮9\)

\(\Rightarrow\left(12+x+y\right)⋮9\)

Do \(0\le x,y\le9\)

\(\Rightarrow\left[{}\begin{matrix}x+y=6\\x+y=15\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;5\right),\left(5;1\right),\left(2;4\right),\left(4;2\right),\left(3;3\right),\left(6;9\right),\left(9;6\right),\left(8;7\right),\left(7;8\right)\right\}\)

\(35-5\left(x-1\right)=10\\ \Leftrightarrow35-5x+5=10\\ \Rightarrow40-5x=10\)

\(\Rightarrow-5x=10-40\\ \Rightarrow-5x=-30\\ \Rightarrow x=\dfrac{-30}{-5}=6\)

c) 

\(24\left(x-16\right)=12^2\)

\(\Rightarrow24x-384=144\\ \Rightarrow24x=144+384\\ \Rightarrow24x=528\\ \Rightarrow x=\dfrac{528}{24}=22\)

d) 

\(\left(x^2-10\right)\div5=3\\ \Rightarrow\left(x^2-10\right)=3\times5\\ \Rightarrow x^2-10=15\)

\(\Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\Rightarrow x=5\)

Ơ, bn ơi @Kenny sai sai ở đâu thì phải.

Ta có : 

\(\left|3x+18\right|\ge0\) và \(\left|4x-28\right|\ge0\) \(\Rightarrow\) \(\left|3x+18\right|+\left|4y-28\right|\ge0\)

Mà \(\left|3x+18\right|+\left|4y-28\right|\le0\) ( đề bài cho )

\(\Rightarrow\)\(\left|3x+18\right|+\left|4y-28\right|=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+18=0\\4y-28=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-18\\4y=28\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\y=7\end{cases}}}\)

Vậy \(x=-6\) và \(y=7\)

Ta có \(\left|3x+18\right|+\left|4y-28\right|\le0\)

Mà \(\left|3x+18\right|\ge0\forall x;\left|4y-28\right|\ge0\forall y\)

=> |3x+18|+|4y-28|=0

=> 3x+18=4y-28=0

• 3x+18=0 <=> 3x=-18 <=> x=-6

• 4y-28=0 <=> 4y=28 <=> y=7

Vậy ...

thank you very much

`4x^2=13`

`<=>x^2=13/4`

`<=>x=[+-\sqrt{13}]/2`

Vậy `S={[+-\sqrt{13}]/2}`

`4x^2 = 13`.

`=> x^2 = 13/4`.

`=> x = (sqrt 13)/(sqrt 4)`

`=> x = (+-sqrt 13)/2`.

Vậy `S = (+-sqrt 13)/2`.

\(4x^2=13\)

\(x^2=52\)

\(x=\sqrt{52}\)

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)